UNIVERSIDADE DE SÃO PAULO
Instituto de Ciências Matemáticas e de Computação
ISSN 0103-2577
_______________________________
PROPERTIES OF SOLUTIONS OF A CLASS OF HYPOCOMPLEX VECTOR FIELDS
CAMILO CAMPANA P. L. DATTORI DA SILVA
A. MEZIANI
No 417
_______________________________
NOTAS DO ICMC
SÉRIE MATEMÁTICA
São Carlos – SP
Properties of solutions of a class of hypocomplex vector fields
Camilo Campana
∗, Paulo L. Dattori da Silva
†and A. Meziani
Abstract
A Cauchy type integral operator is associated to a class of integrable vector fields with complex coefficients. Properties of the integral operator are used to deduce H¨older solvability of semilinear equations Lu = F(x, y, u) and a strong similarity principle between the solutions of the equation Lu = au+bu and those of the equation Lu= 0.
1 Introduction
This paper explores the extent to which properties of the Cauchy-Riemann operator
∂/∂z extend to planar complex vector fields. The class of vector fields amenable to possess such properties is within the so called “hypocomplex” vector fields. In the plane these vector fields have first integrals that are local homeomorphisms. Any such vector field L with C∞ coefficients is solvable in the C∞ category: if f ∈ C∞(U) then there exists u∈C∞(U) satisfying the equation
Lu=f. (1)
In fact, L is also hypoelliptic. It is also proved (see [1] or [5]) that if f ∈ Lp then (1) has solutions in Lp. However, in general the solution u cannot be expected to be more regular than the right hand side. Indeed, in [1] the authors give an example of a C∞ hypocomplex vector field L and a functionf ∈L∞(R2) such that (1) does not haveL∞ solutions in any neighborhood of 0∈R2.
In this paper we isolate a class of hypocomplex vector fields, with properties analogous to those of∂/∂z. This class consist of those vector fields L, defined in a bounded domain Ω, that are locally equivalent to a multiple of the vector field
∂/∂y−i|y|σp∂/∂x, σp >0, (2)
∗Supported by FAPESP (grants 2013/26463-7 and 2013/08452-8).
†Partially supported by CNPq (grants 306076/2012-8 and 478542/2013-5) and FAPESP (grants 2012/03168-7 and 2014/06515-5).
in a neighborhood of each point p whereL fails to be elliptic. In this case, we prove that if f ∈Lp(Ω), with p >2 +σ and σ= max{σp}, then all solutions u of (1) are in Cα(Ω), where α = (2−q−τ)/q, with τ =σ/(σ+ 1) and q =p/(p−1). This result is obtained through the study of the integral operator
TZf(x, y) = 1 2πi
Z
Ω
f(ξ, η)
Z(ξ, η)−Z(x, y)dξdη, (3)
where Z : Ω → C is a global first integral of L. This result is then applied to show that the semilinear equation
Lu=F(x, y, u) (4)
has H¨older continuous solution for a class of functions F. As a consequence we obtain a strong similarity principle for the solutions of the equation
Lu=Au+Bu, (5)
with A, B ∈ Lp(Ω). That is, any solution of (5) is in Cα(Ω) and satisfies u = H(Z)es, where H is a holomorphic function defined in Z(Ω) and s∈Cα.
It should be noted that analogous questions were investigated in [10] for vector field (2) with σ ∈ 2Z+ and where f ∈ L∞ (resp, A, B ∈ L∞). The approach and motivation for this paper are related to [1], [2], [3], [4], [7], [8], [9], [10], [13], and many others. This paper is organized as follows. After the necessary preliminaries of section 2 and technical Lemmas of section 3, we study properties of the integral operator TZ in section 4 and 5. In particular, we prove that TZf solves (1) for any f ∈ L1(Ω), TZf ∈ Lq(Ω) for any 1 ≤ q < 2−τ, and TZf ∈ C(2−q−τ)/q(Ω) if f ∈ Lp(Ω) with p > 2 +σ. In section 6 we study the semilinear equation and deduce the similarity principle.
This work was done when the first and second authors were visiting the Department of Mathematics & Statistics at Florida International University. They are grateful and would like to thank the members of the host institution for the support they provided during the visit.
2 Preliminaries
Let
L=A(x, y)∂/∂x+B(x, y)∂/∂y
be a complex vector field defined in a region ˜Ω⊂R2, whereAand B areC-valued H¨older continuous functions in ˜Ω, |A|+|B| > 0 in ˜Ω. Let Ω be an open set such that Ω ⊂ Ω.˜ Hypocomplex vector fields were introduced (see [1] or [12]) as those vector fields that are locally integrable and such that any solution ofLu= 0 can be written locally ash◦Z with h holomorphic and Z a first integral. It turns out that in the case of vector fields in the
plane, this is equivalent to requiring that any first integral ofLis a local homeomorphism.
In this paper, we consider a vector field Lto behypocomplex on Ω if for everyp∈Ω there exists a C1+ function (for some >0) Z :U →C defined in an open setU, p∈ U, such that dZ 6= 0, LZ = 0, and Z :U → Z(U) is a homeomorphism. The set where L fails to be elliptic is given by
Σ ={p∈Ω; Lp∧Lp= 0}={p∈Ω; Im(AB)(p) = 0},
where L=A(x, y)∂/∂x+B(x, y)∂/∂y is the complex conjugate of L. We will refer to Σ as the characteristic set of L (Σ is in fact the base projection of the characteristic set of the first order differential operator L).
The class of vector fields under study in this paper are those vector fields that satisfy the following conditions:
(i) L is hypocomplex in Ω
(ii) The characteristic set Σ⊂Ω is a C1+ curve nontangent to L
(iii) For every p∈Σ, there exists an open setU, with p∈ U, such that Σ∩ U is given by a defining C1+ function ρ(x, y) such that =(AB)(x, y) =|ρ(x, y)|σg(x, y), for some continuous function g inU satisfying g(x, y)6= 0 for all (x, y)∈ U.
It should be noted that these conditions are invariant underC1+ change of variables.
The following Proposition gives a local normal form for vector fields satisfying the above conditions.
Proposition 2.1. Suppose that L satisfies (i), (ii), and (iii). Then, for every p ∈ Σ, there exist an open neighborhoodU such thatU\Σ consists of two connected components U+ and U−, and local coordinates (x+, t+) (respectively (x−, t−)) centered at psuch that L is a multiple of the following vector field in U+ (respectively U−):
Lσ =∂/∂t±−i|t±|σ∂/∂x±, (6) with first integral
Zσ(x, t) = x±+it±|t±|σ
σ+ 1 (7)
Proof: Since L is hypocomplex and Σ is smooth then we can assume that there is in a neighborhood of p ∈ Σ, such that the set Σ∩ U is given by {y = 0} and that the first integral has the form Z(x, y) =x+iϕ(x, y), with ϕ real-valued (see [1], [12]). Thus L is a multiple of the Hamiltonian
Zx∂/∂y−Zy∂/∂x and condition (iii) implies that
∂ϕ
∂y(x, y) = |y|σψ(x, y),
with ψ a C0 function, and ψ(0,0)6= 0. Then, ϕ(x, y) =
Z y
0
|s|σψ(x, s)ds +β(x) = ˜ϕ(x, y) +β(x), for some C1+-function β. With respect to the new variables
x=x, t= sgn( ˜ϕ)|ϕ(x, y)(1 +˜ σ)|1+σ1 the expression of Z becomes
Z(x, t) = x+i
t|t|σ
σ+ 1 +β(x)
.
Let D be a small disc centered at 0 such that Σ divides D into two semi discs D+ = {t > 0} and D− = {t < 0}. Note that Z(D+) and Z(D−) are simply connected in C sharing a boundary curve γ = {(x, β(x))}. Since γ is C1+-curve, then we can find conformal mappings
H±: Z(D±) −→ H±(Z(D±))⊂C
sending the boundary curve γ into the real axis and H± extends as aC1-diffeomorphism to a full neighborhood of 0. The functionZ±(x, t) = H±(Z(x, t)) is then a first integral of L in D± satisfying =Z±(x,0) = 0. It follows that=Z±(x, t) = t|t|σ
σ+ 1
ψ(x, t)˜ 1+σ for some function ˜ψ(x, t) (with ˜ψ(0,0)>0). With respect to the new coordinates
x± =<Z±(x, t), t±=tψ(x, t)˜
the vector fieldLbecomes a multiple of the desired vector fieldLσgiven in the Proposition.
Remark 2.2. Note that if L is C∞, locally integrable and, of constant type along each connected component of Σ, then it satisfies (iii) (see [1], [12]). Indeed, in this case if Σj is connected component of Σ and Lis of constant typenj along Σj, then it can be shown (as in [10]) that for each point p∈Σj coordinates (x±, t±) can found in which the expression of L is as in the Proposition 2.1.
Remark 2.3. The vector field (of infinite type) L= ∂
∂t −ie−|t|1 t2
∂
∂x,
is of class C∞, with characteristic set Σ ={t= 0}. Also, L is hypocomplex with (global) first integral
Z(x, t) =x+i t
|t|e−|t|1 .
However, the condition (iii) is not satisfied, since e−|t|1 6=O(|t|σ), for any σ >0.
We close this section with the following Proposition.
Proposition 2.4. Let L be a hypocomplex vector field with local first integrals defined in an open set ˜Ω ⊂R2. Then for any open set Ω ⊂⊂ Ω,˜ L has a global first integral on Ω. More precisely, there exists C1+ function
Z : Ω −→ Z(Ω)⊂⊂C such that Z is a homeomorphism, LZ = 0 and dZ 6= 0.
Proof: Since L is hypocomplex, then for every p ∈ Ω there exists a C1+-first integral Zp defined in an open set Up ⊂ Ω with˜ p ∈ Up. The collection {Up, Zp}p∈Ω defines a structure of a Riemann surface on the open set S
p∈ΩUp, since the transition functions hpq = Zp ◦Zq−1 are holomorphic functions on Zq(Up ∩Uq). The existence of a global first integral Z follows from the Uniformization Theorem of the planar Riemann surfaces S
p∈ΩUp (see [11]).
Remark 2.5. Note that hypocomplexity of Lalso implies that ifv satisfiesLv = 0 in an open set O ⊂Ω, then v =h◦Z, whereh is a holomorphic function defined in Z(O).
3 Some Lemmas
We prove some technical Lemmas that will be used in the following sections.
Lemma 3.1 ([10], Lemma 3.1). Let 0 < δ < R, 0 < τ < 1, m > 0, and 0 ≤ γ < R.
Then, there exists a constant C(τ)>0 such that Z R
δ
Z 2π
0
dθdr
|γ+rsinθ|τrm+1 ≤ C(τ) δm+τ.
Lemma 3.2. Let R > 0, 0 < τ < 1, γ ∈ R, and 1 < q < 2−τ. Then, there exists a constant M(q, τ)>0 such that
I = Z 2π
0
Z R
0
drdθ
|γ+rsinθ|τrq−1 ≤M(q, τ)R2−τ−q. Proof: To prove the Lemma it is enough consider γ ≥0. If γ = 0 then
I = Z 2π
0
Z R
0
drdθ
|rsinθ|τrq−1 = Z 2π
0
dθ
|sinθ|τ Z R
0
dr
rτ+q−1 =C(τ) R2−τ−q 2−τ −q, where
C(τ) = Z 2π
0
dθ
|sinθ|τ <∞.
Now, suppose γ >0. Let r =ργ. We have I =γ2−τ−q
Z 2π
0
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1
=. γ2−τ−qJ. (8)
To estimate J, we consider two cases: 0< γ < R and γ ≥R.
Assume that 0< γ < R. We can write J =
Z π
0
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1 + Z 2π
π
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1
=. J1+J2.
Forθ∈[0, π] we have sinθ≥0; consequently, ρsinθ+ 1≥ρsinθ andρ∈[0, R/γ]. Hence, J1 ≤
Z π
0
Z Rγ
0
dρdθ
|ρsinθ|τρq−1 ≤ C(τ) 2−τ −q
R γ
2−τ−q .
Next, we will estimate J2. Let π < θ0 <3π/2 such that −sinθ0 =γ/2R. We can write J2 =
Z θ0
π
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1 + Z 3π2
θ0
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1 +
Z 3π−θ0
3π 2
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1 + Z 2π
3π−θ0
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1
=. J2,1+J2,2+J2,3+J2,4. Let ϕ=−θ+ 3π. We have
J2,3 =
Z 3π−θ0
3π 2
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1 = Z 3π2
θ0
Z Rγ
0
dρdϕ
|ρsinϕ+ 1|τρq−1 =J2,2. Also,
J2,4 = Z 2π
3π−θ0
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1 = Z θ0
π
Z Rγ
0
dρdϕ
|ρsinϕ+ 1|τρq−1 =J2,1. Let us estimate J2,1. Note that
0≤ −sinθ ≤ γ 2R < 1
2, for 0∈[π, θ0];
consequently, for θ∈[π, θ0] and 0< ρ < R/γ, we have ρ|sinθ|< 1
2 <1 +ρsinθ.
Hence,
J2,1 ≤ Z θ0
π
Z Rγ
0
dρdθ
ρτ|sinθ|τρq−1 ≤ C(τ) 4(2−τ −q)
R γ
2−τ−q
.
Let us estimate J2,2. Let ϕ=θ and t=ρsinθ+ 1. We have J2,2 =
Z 3π2
θ0
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1 = Z 3π2
θ0
Z Rγsinϕ+1
1
1 sinϕ
dtdϕ
|t|τ
t−1 sinϕ
q−1
= Z 3π2
θ0
|sinϕ|q−2Z 1
R γ sinϕ+1
dt
|t|τ|t−1|q−1
dϕ
≤ Z 3π2
θ0
|sinϕ|q−2Z 1
−Rγ
dt
|t|τ|t−1|q−1
dϕ.≤C(q, τ) R
γ
2−τ−q
, for some constant C(q, τ)>0.
Hence, by the calculations above, we can find a constantC1(q, τ)>0 such that J ≤C1(q, τ)
R γ
2−τ−q
. (9)
Now, assume thatγ ≥R. In this case, there exists a constant C(q, τ)>0 such that J=
Z 2π
0
Z Rγ
0
dρdθ
|ρsinθ+ 1|τρq−1 ≤ Z 2π
0
Z Rγ
0
dρdθ
|1−ρ|τρq−1 ≤C(q, τ) R
γ
2−τ−q
(10) Finally, estimates (8), (9) and (10) show that there exists a constantM(q, τ)>0 such that
I ≤M(q, τ)R2−τ−q.
Lemma 3.3. Let R >0, γ ≥0, 0< τ <1, 1≤ q < 2−τ, and 0 ≤ϕ≤π. Then, there exists a constant C(q, τ)>0 such that
H = Z 2π
0
Z R
0
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ −1|q ≤C(q, τ). (11) Proof: We will divide the proof in two cases: γ = 0 andγ >0.
Case 1: γ = 0. In this case we can write H =
Z 2π
0
Z R
0
drdθ
|sin(θ+ϕ)|τrq+τ−1|reiθ −1|q =H1+H2+H3,
with
H1 = Z 2π
0
Z 12
0
drdθ
|sin(θ+ϕ)|τrq+τ−1|reiθ−1|q, H2 =
Z 2π
0
Z 32
1 2
drdθ
|sin(θ+ϕ)|τrq+τ−1|reiθ−1|q, H3 =
Z 2π
0
Z R
3 2
drdθ
|sin(θ+ϕ)|τrq+τ−1|reiθ −1|q. For 0< r < 1
2 we have |reiθ−1| ≥ 1
2. Hence, we can find C1(q, τ)>0 such that H1 ≤2q
Z 2π
0
Z 12
0
drdθ
|sin(θ+ϕ)|τrq+τ−1 ≤C1(q, τ).
For 1
2 < r < 3
2 we have 2
3 τ+q
< 1
rτ+q <2τ+q, and so H2 ≤2τ+q
Z 2π
0
Z 32
1 2
drdθ
|sin(θ+ϕ)|τ|reiθ −1|q
= 2τ+q
4
X
j=1
Z jπ2
(j−1) 2
Z 32
1 2
drdθ
|sin(θ+ϕ)|τ|reiθ−1|q
= 2. τ+q
4
X
j=1
H2,j.
Let 0 < θ0 < π/2 such that 1−θ2
2 <cosθ <1− θ2
4 , for 0< θ < θ0. Then
H2,1= Z θ0
0
Z 32
1 2
drdθ
|sin(θ+ϕ)|τ|reiθ−1|q + Z π2
θ0
Z 32
1 2
drdθ
|sin(θ+ϕ)|τ|reiθ −1|q=H2,11 +H2,12 . For 0 < θ < θ0 and 1/2≤r ≤3/2 we have
(r−1)2+θ2
4 ≤(r−1)2+ θ2
4 ≤(r−1)2+rθ2
2 ≤ |reiθ−1|2 and, consequently,
1
|reiθ−1|2 ≤ 4
(r−1)2+θ2.
Hence,
H2,11 ≤2q Z θ0
0
Z 32
1 2
drdθ
|sin(θ+ϕ)|τ[(r−1)2+θ2]q2
= 2. qJ.
Suppose that 0≤ϕ≤π/2. Then, 0< θ+ϕ < π/2 +θ0 < π and|sin(θ+ϕ)| ≥C|θ+ϕ|, for some constant C > 0. Hence,
CτJ ≤ Z θ0
0
Z 32
1 2
drdθ
|θ+ϕ|τ[(r−1)2+θ2]q2 ≤ Z
D((0,1);1)
drdθ
|θ+ϕ|τ[(r−1)2+θ2]q2
= Z 1
0
Z 2π
0
dρdt
|ϕ+ρsint|τρq−1 ≤M(q, τ),
where the last inequality is obtained by Lemma 3.2. Similar estimate can obtained in the case π/2< ϕ≤π. Therefore,
Z θ0
0
Z 32
1 2
drdθ
|sin(θ+ϕ)|τ|reiθ−1|q ≤ 2qM(q, τ)
Cτ . (12)
Forθ0 < θ < π2 we have
|reiθ −1|2 ≥1−cosθ≥1−cosθ0 and
H2,11 = Z π2
θ0
Z 32
1 2
drdθ
|sin(θ+ϕ)|τ|reiθ−1|q
≤ 1
(1−cosθ0)q2 Z π2
θ0
dθ
|sin(θ+ϕ)|τ ≤ C1(τ) (1−cosθ0)q2,
(13)
for some constant C1(τ)>0. It follows from (12) and (13), that H2,1 ≤ 2qM(q, τ)
Cτ + C1(τ)
(1−cosθ0)q2. (14)
To estimate H2,2, we start by using a change of variable θ =α+π in the integral to obtain
H2,2 = Z 2π
3π 2
Z 32
1 2
drdα
|sin(α+ϕ)|τ|reiα+ 1|q. Note that |reiα + 1| ≥rcosα+ 1 ≥1, for 3π/2≤α≤2π; hence,
H2,2 ≤ Z 2π
3π 2
dα
|sin(α+ϕ)|τ =C2(τ)<∞. (15)
Similar estimates can be obtained for H2,3 and H2,4. Therefore, we can find a constant C2(q, τ)>0 such that H2 ≤C2(q, τ).
The estimation of H3 is obtained as follows. Sincer >3/2, then |reiθ−1| ≥ |r−1|= r−1≥r/3 and
H3 ≤3q Z 2π
0
Z R
3 2
drdθ
|sin(θ+ϕ)|τr2q+τ−1
≤3qC3(τ) Z R
3 2
dr
r2q+τ−1 ≤ 3qC3(τ) 2q+τ−2
2 3
2q+τ−2
,
where C3(τ) is a positive constant. This completes the proof of (11) in the case γ = 0.
Case 2: γ >0. In this case,H (given by (11)) can be rewritten in the formH =H1+H2, where
H1 = Z 2π
0
Z γ
0
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ −1|q, H2 =
Z 2π
0
Z R
γ
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ−1|q. To estimate H1, we consider three cases depending on the values of γ.
First case: 0< γ ≤ 1
2. In this case, since 0< r <1/2, we have |reiθ−1| ≥1−r≥ 1/2.
Hence, by Lemma 3.2, H1 ≤2q
Z 2π
0
Z γ
0
drdθ
|γ+rsin(θ+ϕ)|τrq−1 ≤2qM(q, τ).
Second case: 1
2 < γ≤ 3
2. We writeH1 =H1,1+H1,2, where H1,1 =
Z 2π
0
Z 12
0
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ −1|q, H1,2 =
Z 2π
0
Z γ
1 2
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ−1|q. Note that estimation ofH1,1 is given by the previous case.
ForH1,2 we have H1,2 ≤2q−1
Z 2π
0
Z γ
1 2
drdθ
|γ−r|τ|reiθ−1|q
= 2q−1
4
X
j=1
Z jπ2
(j−1)π 2
Z γ
1 2
drdθ
|γ−r|τ|reiθ−1|q = 2q−1
4
X
j=1
H1,2j .
To estimate H1,21 we write H1,21 =
Z θ0
0
Z γ
1 2
drdθ
|γ−r|τ|reiθ −1|q + Z π2
θ0
Z γ
1 2
drdθ
|γ−r|τ|reiθ−1|q,
with 0 < θ0 < π/2 such that 1−θ2/2 <cosθ < 1−θ2/4 for 0< θ < θ0. Note that this choice of θ0 implies that|reiθ−1|>1−cosθ0 for θ≥θ0. Hence,
H1,21 ≤2q Z θ0
0
Z γ
1 2
drdθ
|γ−r|τ((r−1)2+θ2)q2 +
π 2 −θ0 (1−cosθ0)q2
1 1−τ.
Now, by using polar coordinates r = 1 +ρsinφ, θ = ρcosφ in the integral, and using Lemma 3.2, we obtain
Z θ0
0
Z γ
1 2
drdθ
|γ−r|τ((r−1)2+θ2)q2 ≤ Z 1
0
Z 2π
0
dϕdρ
|γ−1−ρsinϕ|τρq−1 ≤M(q, τ).
Hence, we can find a constant C1(q, τ) > 0 such that H1,21 ≤ C1(q, τ). Proceeding as above, we can find a constant C2(q, τ)>0 such that H1,24 ≤C2(q, τ).
ForH1,22 , we have H1,22 =
Z 2π
3π 2
Z γ
1 2
drdθ
|γ−r|τ|reiθ+ 1|q ≤ Z 2π
3π 2
Z γ
1 2
drdθ
|γ−r|τ ≤ π 2(1−τ),
since |reiθ + 1| ≥ 1 for 3π2 ≤ θ ≤ 2π. Analogous estimates can be used to show that H1,23 ≤ π
2(1−τ). Therefore, we can find a constant C3(q, τ)>0 such that H1 ≤C3(q, τ).
Third case: γ > 3
2. In this case we write H1 =H1,1+H1,2, where H1,1 =
Z 2π
0
Z 32
0
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ −1|q, H1,2 =
Z 2π
0
Z γ
3 2
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ−1|q.
Estimate for H1,1 follows from the previous case. For H1,2, note that since r ≥ 3/2 we have |reiθ −1| ≥r/2 and
H1,2 ≤2q Z 2π
0
Z γ
3 2
drdθ
|γ+rsin(θ+ϕ)|τr2q−1 =C(q, τ)<∞.
Now we estimate H2. Recall that H2 =
Z 2π
0
Z R
γ
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ −1|q. As done for H1 we consider three cases:
First case: 3
2 ≤γ < R. Since r > 3
2 implies|reiθ−1| ≥ r
3, we have H2 ≤3q
Z 2π
0
Z R
γ
drdθ
|γ+rsin(θ+ϕ)|τr2q−1 = 3q Z 2π
0
Z R
γ
drdθ
|γ+rsinθ|τr2q−1. Hence, the estimative for H2 follows from Lemma 3.1.
Second case: 1
2 ≤γ < 3
2. In this case we can write H2 =H2,1+H2,2, where H2,1 =
Z 3π2
−π2
Z 32
γ
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ −1|q =H2,11 +H2,12 , H2,2 =
Z 2π
0
Z R
3 2
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ−1|q and,
H2,11 = Z π2
−π
2
Z 32
γ
drdθ
|γ+rsinθ|τrq−1|rei(θ−ϕ)−1|q, H2,12 =
Z 3π2
π 2
Z 32
γ
drdθ
|γ +rsinθ|τrq−1|rei(θ−ϕ)−1|q. Note that estimation of H2,2 is given by the previous case.
To estimate H2,11 , note that since 1/2 ≤ γ < r < 3/2 we have 1/3 < γ/r < 1. Let θ1 ∈(−π/2,−α0) be such that −γ/r= sinθ1, where sinα0 = 1/3. We have then
1
|γ+rsinθ|τ = |θ−θ1|τ rτ|sinθ−sinθ1|τ
1
|θ−θ1|τ ≤ A(τ) rτ|cosθ1|τ
1
|θ−θ1|τ, for some constant A(τ)>0; hence, substituting cosθ1 byp
r2 −γ2/r we obtain 1
|γ+rsinθ|τ ≤ A(τ) (r2−γ2)τ2
1
|θ−θ1(r)|τ.
Therefore,
H2,11 ≤2q−1A(τ) Z π2
−π
2
Z 32
γ
drdθ
(r2−γ2)τ2|θ−θ1(r)|τ|rei(θ−ϕ)−1|q
= 2q−1A(τ) Z π2−ϕ
−π
2−ϕ
Z 32
γ
drdθ
(r2−γ2)τ2|θ−θ1(r) +ϕ|τ|reiθ −1|q
≤2q−1A(τ)
4
X
j=1
Z (j−3)π2
(j−4)π 2
Z 32
γ
drdθ
(r2−γ2)τ2|θ−θ1(r) +ϕ|τ|reiθ −1|q
= 2q−1A(τ)
4
X
j=1
H2,11,j. We have
H2,11,1 = Z 0
−π
2
Z 32
γ
drdθ
(r2−γ2)τ2|θ−π−θ1(r) +ϕ|τ|reiθ + 1|q. For −π/2≤θ ≤0 we have |reiθ+ 1| ≥1 +rcosθ ≥1 and
H2,11,1 ≤ Z 0
−π2
Z 32
γ
drdθ
(r2−γ2)τ2|θ−π−θ1(r) +ϕ|τ ≤C(τ), for some constant C(τ)>0. Similarly H2,11,2 ≤C(τ˜ ), for some ˜C(τ)>0.
To estimate H2,11,3, let 0< θ0 < α0/2 such that 1−θ2
2 <cosθ <1− θ2
4, for −θ0 ≤θ ≤0.
Note that −θ0 ≤θ ≤0 implies
θ−θ1(r) +ϕ≥ −θ0−θ1(r) +ϕ≥α0− α0
2 +ϕ≥ α0 2 . Moreover, for −π/2≤θ ≤ −θ0 we have
|reiθ−1|2 ≥1−cosθ0. Hence,
H2,11,3 ≤ 2
α0
τZ 0
−θ0
Z 32
γ
drdθ
(r2−γ2)τ2|reiθ−1|q +
1
1−cosθ0
q2 Z −θ0
−π2
Z 32
γ
drdθ
(r2−γ2)τ2|θ−θ1(r) +ϕ|τ
≤ 2
α0
τZ 0
−θ0
Z 32
γ
drdθ
(r2−γ2)τ2|reiθ−1|q +C1(τ, q),
for some constant C1(τ, q)>0. Note that for r >1/2 and −θ0 ≤θ ≤0 we have
|reiθ−1|2 ≥ (r−1)2+θ2
4 ;
hence, since r ≥γ ≥1/2 implies r2−γ2 ≥r−γ, we have Z 0
−θ0
Z 32
γ
drdθ
(r2−γ2)τ2|reiθ−1|q ≤2q Z 0
−θ0
Z 32
γ
drdθ
(r−γ)τ2((r−1)2+θ2)q2
≤2q Z
D((0,1);1)
drdθ
(r−γ)τ2((r−1)2+θ2)q2
= 2q Z 1
0
Z 2π
0
drdθ
|1−γ+ρsinα|τ2ρq−1 ≤C2(τ, q),
where the last estimate follows from Lemma 3.2. Therefore, we can find a constant C3(τ, q) >0 such that H2,11,3 ≤ C3(τ, q). Similar arguments can be used to estimate H2,11,4 and H2,12 . Therefore, we can find a constant C(τ, q)>0 for which H2,1 ≤C(τ, q).
Third case: 0< γ ≤ 1
2. In this case H2=
Z 2π
0
Z 12
γ
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ−1|q + Z 2π
0
Z R
1 2
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ −1|q. The second integral can be estimate as before. Hence, it is enough to estimate the first integral. Since r <1/2 implies |reiθ −1| ≥ |r−1| ≥1/2, we have
Z 2π
0
Z 12
γ
rdrdθ
|γ+rsin(θ+ϕ)|τrq|reiθ−1|q ≤2q Z 2π
0
Z 12
0
drdθ
|γ+rsin(θ+ϕ)|τrq−1, which can be estimated applying Lemma 3.2.
This completes the proof of the caseγ >0 and of the Lemma.
4 An Integral Operator
Let L be a vector field defined in an open set ˜Ω ⊂ R2 and Ω be a relatively compact subset of ˜Ω (Ω⊂ Ω) with˜ ∂Ω piecewise of class C1. Suppose that L satisfies conditions (i), (ii), and (iii). Let
Z : Ω→Z(Ω)⊂C (16)
be a global first integral of L of classC1+ as in Proposition 4. As noted in section 2, L is a multiple of the Hamiltonian of Z. In fact, we will assume that
L=Zx ∂
∂y −Zy ∂
∂x. (17)
For f ∈L1(Ω), define the integral operator TZf(x, y) = 1
2πi Z
Ω
f(ξ, η)
Z(ξ, η)−Z(x, y)dξdη. (18)
Since Lsatisfies condition (iii), then for every Σj, connected component of Σ, there exists σj >0 such that (7) holds. Let
σ = maxσj. (19)
We have the following theorem:
Theorem 4.1. Let σ be given by (19) and let f ∈ Lp(Ω), with p > 2 +σ. Then, there exists a constant M(p, σ,Ω)>0 such that the operator TZ satisfies
|TZf(x, y)| ≤M(p, σ,Ω)||f||p, ∀(x, y)∈Ω.
Proof: Let (x, y)∈Ω. Then
|TZf(x, y)| ≤ 1 2π
Z
Ω
|f(ξ, η)|
|Z(ξ, η)−Z(x, y)|dξdη ≤ ||f||p 2π
Z
Ω
dξdη
|Z(ξ, η)−Z(x, y)|q 1q
where q = p/(p−1). Since Z is a C1+-diffeomorphism outside of the characteristic set Σ, then it follows from the normalization given in Proposition 2.1 and use of partition of unity that in order to have
Z
Ω
dξdη
|Z(ξ, η)−Z(x, y)|q ≤M(p, σ,Ω)
it is enough to have the inequality when Ω is replaced by Ω∩ U, when U is a open set where the normalization holds. For the purpose of estimating the integral, there is no loss of generality in assuming that the vector field can be transformed into the normal form Lσ in the open set U rather than in each connected component U+ andU− of U \Σ (since the diffeomorphisms extend across the boundary by Proposition 2.1).
Letw∈Σj ⊂Σ. There exists a C1-diffeomorphism
Φ :D(0, δ)→Uw = Φ(D(0, δ))⊂R2
with Φ(0) = w, and a holomorphic function H defined on Z(Uw), with H0(ζ) 6= 0 for ζ ∈Z(Uw), such that
Z◦Φ(s, t) =H(Zj(s, t)), where Zj(s, t) = s+i t|t|σj
1 +σj
. Hence for (x, y)∈Uw
Z
Ω∩Uw
dξdη
|Z(ξ, η)−Z(x, y)|q ≤ Z
D(0,δ)
|DΦ(s, t)|dsdt
|H(Zj(s, t))−H(Zj(s◦, t◦))|q
≤ max|DΦ(s, t)|
min|H0(ζ)|q Z
D(0,δ)
dsdt
|Zj(s, t)−Zj(s◦, t◦)|q. By using the change of variables ξ =s and η = t|t|σj
1 +σj we obtain (1 +σj)τj
Z
D(0,δ)
dsdt
|Zj(s, t)−Zj(s◦, t◦)|q = Z
Zj(D(0,δ))
dξdη
|η|τ|ζ−z|q
≤ Z
D(z,d)
dξdη
|η|τ|ζ−z|q, where ζ =ξ+iη, z =Zj(s◦, t◦), andd= diam(Zj(D(0, δ))).
Now, the use of polar coordinates (r, θ) given byξ =rcosθ+<(z) andη=rsinθ+=(z) and of Lemma 3.3 give
Z
D(z,d)
dξdη
|η|τ|ζ−z|q = Z 2π
0
Z d
0
drdθ
|rsinθ+=(z)|τrq−1 ≤M(q, τ)d2−τ−q.
Proposition 4.2. Letσ be given by(19) and let f ∈L1(Ω). Then,T f ∈Lq(Ω), for any 1≤q <2−σ/(σ+ 1).
Proof: Let f ∈ L1(Ω) and let g ∈ Lp(Ω), with p > 2 +σ. It follows from Lemma 3.2 that the function
g1(x, y) = Z
Ω
|g(ξ, η)| dξdη
|Z(ξ, η)−Z(x, y)|
is bounded. Hence, f g1 ∈L1(Ω). By applying Fubini’s Theorem we obtain Z
Ω
|f(x, y)|g1(x, y)dxdy= Z
Ω
|f(x, y)|
Z
Ω
|g(ξ, η)| dξdη
|Z(ξ, η)−Z(x, y)|
dxdy
= Z
Ω
|g(ξ, η)|
Z
Ω
|f(x, y)| dxdy
|Z(ξ, η)−Z(x, y)|
dξdη
= Z
Ω
|g(ξ, η)|f1(ξ, η)dξdη, where
f1(ξ, η) = Z
Ω
|f(x, y)| dxdy
|Z(x, y)−Z(ξ, η)|.
Hence |g|f1 ∈ L1(Ω). Since g ∈ Lp(Ω) is arbitrary, it follows from the converse of the H¨older inequality (see, for instance, [14]; also, [6]) that f1 ∈ Lq(Ω), for q = p/(p−1).
Note that p > 2 +σ implies 1< q <2−σ/(σ+ 1). Also, note that |T f| ≤f1. Therefore,
T f ∈Lq(Ω), for any 1< q <2−σ/(σ+ 1).
The following Lemma is a direct consequence of Green’s Theorem.
Lemma 4.3. Let w∈C(Ω)∩C1(Ω). Then Z
Ω
Lw dxdy =− Z
∂Ω
w dZ(x, y). (20)
Proposition 4.4. Let w∈C(Ω)∩C1(Ω). Then, for all (x, y)∈Ω, we have w(x, y) = 1
2πi Z
∂Ω
w(α, β)
Z(α, β)−Z(x, y)dZ(α, β) + 1 2πi
Z
Ω
Lw(α, β)
Z(α, β)−Z(x, y) dαdβ.
Proof: Let (x0, y0) ∈ Ω fixed. Set z0 = Z(x0, y0) and let > 0 such that D ⊂ Z(Ω), where D =D(z0, ). Define K =Z−1(D) and Ω = Ω\K. Let
f(x, y) = w(x, y) Z(x, y)−z0. We have f ∈C(Ω)∩C1(Ω). Hence, by (20),
Z
Ω
Lw(x, y)
Z(x, y)−z0 dxdy=− Z
∂Ω
w(x, y)
Z(x, y)−z0dZ(x, y)
=− Z
∂Ω
w(x, y)
Z(x, y)−z0dZ(x, y) + Z
∂K
w(x, y)
Z(x, y)−z0dZ(x, y);
(21)
Since
Z
∂K
f(x, y)dZ(x, y) = Z
Z(∂K)
(f ◦Z−1)(ζ)dζ = Z
∂B
˜ w(ζ) ζ−z0dζ,
where ˜w=w◦Z−1, then by using polar coordinates ζ =z0+eiθ, θ∈[0,2π], we obtain Z
∂D(z0)
˜ w(ζ) ζ−z0dζ =
Z 2π
0
˜
w(z0+eiθ)
eiθ ieiθdθ→2πiw(z˜ 0), as →0.
Therefore,
Z
∂K
w(x, y) Z(x, y)−z0
dZ(x, y)→2πiw(x0, y0), as →0. (22) On the other hand, as done in the proof of Theorem 4.1,
(x, y)7→ 1
Z(x, y)−z0 ∈Lq(Ω) ⊂L1(Ω).
Hence,
Z
Ω
Lw(x, y)
Z(x, y)−z0 dxdy → Z
Ω
Lw(x, y)
Z(x, y)−z0 dxdy, as →0. (23)
The Proposition follows from (21), (22) and (23).
Theorem 4.5. If f ∈L1(Ω) then TZf satisfies L(TZf) =f in Ω.
Proof: Letf ∈L1(Ω). By Proposition 4.2,TZf ∈Lq(Ω), 1< q <2−σ/(σ+ 1). Hence, by applying Proposition 4.4 we have, for φ ∈C0∞(Ω)
hL(TZf), φi=− Z
Ω
TZf(x, y)Lφ(x, y)dxdy
=− 1 2πi
Z
Ω
Z
Ω
f(ξ, η)
Z(ξ, η)−Z(x, y) dξdη
Lφ(x, y)dxdy
= Z
Ω
f(ξ, η)
− 1 2πi
Z
Ω
Lφ(x, y)
Z(x, y)−Z(ξ, η) dxdy
dξdη
= Z
Ω
f(ξ, η)φ(ξ, η)dξdη =hf, φi.
Therefore, L(TZf) =f in Ω.
Example 4.6. For p ≤ 2 +σ, there exist f ∈ Lp(Ω) such that equation Lu = f has no bounded solutions. Indeed, for the standard vector field L = ∂t−i|t|σ∂x, the function v : Ω→C defined by v(x, t) = ln|ln|Z(x, t)||is not bounded but solves Lv =f with
f(x, t) = −i|t|σ
Z(x, t) ln|Z(x, t)| ∈Lp(Ω), for any 1≤p≤2 +σ.
That f ∈ Lp with p ≤ 2 +σ follows from the fact that Z a
0
dr
rs|lnr|p < ∞ if and only if s ≤ 1. For general vector fields, similar examples can be produced thanks to the normalization near points on Σ.
Remark 4.7. Cauchy type integral operators were used in [8] and [9] in connection with other types of vector fields.
5 H¨ older continuity of solutions
In this section we prove that the solutions of Lu=f are H¨older continuous if f ∈Lp, with p > 2 +σ. Let Σ1,· · · ,ΣN be the connected components of Σ and σ1,· · · , σN be respectively the types of Lalong Σ1,· · · ,ΣN. Recall thatσ = max
1≤j≤N{σj}, τ = σ
σ+ 1, and for p > 2 +σ and q= p
p−1 we have q <2−τ.
Theorem 5.1. Let f ∈ Lp(Ω), with p > 2 +σ. Let q and τ be as given above. If u satisfies Lu=f in Ω, then u∈Cα(Ω) with α= 2−q−τ
q .
Proof: Since L is hypocomplex, it follows from Theorem 4.5 that if u solves Lu=f in Ω, then u=TZf+H(Z), where TZ is the integral operator defined in section 4 andH is a holomorphic function defined onZ(Ω). Thus to prove the Theorem, we need only prove that TZf ∈Cα(Ω).
Let f ∈ Lp(Ω), with p > 2 +σ, and let (x0, y0),(x1, y1) ∈ Ω. Set z1 = Z(x1, y1) and z0 =Z(x0, y0). Then
|TZf(x1, y1)−TZf(x0, y0)| ≤ |z1−z0| 2π
Z
Ω
|f(α, β)|dαdβ
|Z(α, β)−z1||Z(α, β)−z0|
≤ |z1−z0|
2π ||f||pJ1q, where
J = Z
Ω
dαdβ
|Z(α, β)−z1|q|Z(α, β)−z0|q. (24) To prove the Theorem, it is enough to show that
J ≤C1|z1−z0|2−2q−τ+C2, (25)
for some constants C1, C2 >0.
For each j = 1,· · · , N, let Vj be a tubular neighborhood of Σj such that Vj ∩Vk =∅ for j 6=k and such that
Vj =
Mj
[
k=1
Vjk, j = 1,· · · , N,
where each Vjk is an open subset where L can be transformed into the standard vector field Lσj, with first integral
Zj(s, t) = s+i t|t|σj 1 +σj
on each side Vjk± of the characteristic curve Σ∩Vjk (see Proposition 2.1). Hence, we can assume that there exists a diffeomorphisms of class C1
Φ±jk :D±(0, R)→Vjk±
such thatZ◦Φ±jk =Hjk±◦Zj, whereHjk± is a holomorphic function defined inZj(D±(0, R)) and has a C1 extension up to the boundary. Moreover, Hjk0 (ζ)≥Cjk >0 on D±(0, R).
Since L is elliptic outside Σ, then we can use partition of unity to reduce the prob- lem of proving (25) into that of proving the inequality when Ω is replaced by Vjk±, with (x0, y0),(x1, y1)∈Vjk±. In fact for the estimation of the integral, there is no loss of gener- ality in assuming that Lcan be transformed into the standard vector inVjk and not only in each Vjk± separately.
